Timeline for A family of difference sets (paper by A. L. Whiteman)
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 7, 2021 at 7:17 | comment | added | Padraig Ó Catháin | There's a Chapter in Marshall Hall's Combinatorial Theory where all this is laid out in detail. | |
| May 7, 2021 at 7:16 | comment | added | Padraig Ó Catháin | Sorry - I still don't see what you're trying to say. Whiteman's paper is an elaboration of the cyclotomic construction for difference sets. This is normally applied to a group of prime order. It might help you to understand the base case thoroughly, including the relation between the cyclotomic numbers and expressions of the prime as a sum of squares. Then the material in Whiteman's paper is intricate, but nothing one would not expect. In particular, you should see why Whiteman carries out the constructions he does for $-1$, and why this is only a detail in the main argument. | |
| May 7, 2021 at 7:12 | history | edited | Padraig Ó Catháin | CC BY-SA 4.0 |
deleted 21 characters in body
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| May 6, 2021 at 2:19 | comment | added | Akanksha Gupta | This is exactly what my doubt is. I mean I understand the cases when both are odd or either value is even, can verify by examples but not able to prove it for the general case. | |
| May 5, 2021 at 13:38 | comment | added | Padraig Ó Catháin | I am not sure what you are asking. The given value of $m$ solves $(xy)^{m} \equiv -1 \bmod pq$ when both $\frac{p-1}{e}$ and $\frac{q-1}{e}$ are odd. Otherwise, it does not, because there is no solution to the congruence. In that case, you need to multiply by a power of $x$. | |
| May 5, 2021 at 11:31 | comment | added | Akanksha Gupta | So, according to the notations used $m=6$ but I didn't understand the inconsistency I mean what $a$ and $b$. Because in both the cases irrespective of $\frac{p-1}{e}$ and $\frac{q-1}{e}$ even or odd, the order of element $xy$ is lcm(p-1,q-1). | |
| May 5, 2021 at 11:17 | comment | added | Padraig Ó Catháin | I agree that $x \equiv 8 \bmod 35$. And $y \equiv 31 \bmod 35$. Then $xy \equiv 3 \bmod 35$, and this generates a cyclic subgroup of order $12$. But the multiplicative group has order $24$, and the elements which are not powers of $3$ belong to $8\langle 3 \rangle = \{ 8 \cdot 3^j \bmod 35 \}$. You can compute that $8 \cdot 3^9 \equiv -1 \mod 35$. | |
| May 5, 2021 at 10:04 | comment | added | Akanksha Gupta | I tried to look at the case when $pq=35$. The common primitive root for $7$ and $5$ is $3$ and $e=2$, $\frac{5-1}{2}=2$ is even, $\frac{7-1}{2}=3$ is odd. The value of $x$ and $y$ I computed are $8$ and $31$ respectively. From this onward can you elaborate the argument. I'm confused about $a$ and $b$ and the set generated notation. | |
| May 5, 2021 at 9:51 | comment | added | Padraig Ó Catháin | First -- the multiplicative group of $\mathbb{Z}_{pq}$ is not cyclic, so doesn't have a primitive root. Compute some examples to convince yourself of this. Second -- in the third paragraph, $m$ must be even to get a solution $\bmod p$ and odd for a solution $\bmod q$, which is impossible. So no power of $xy$ evaluates to $-1 \bmod pq$. I think your confusion here will be resolved by thinking about my first sentence, and computing some explicit examples. | |
| May 4, 2021 at 21:56 | comment | added | Akanksha Gupta | In the third para, can you elaborate more on the arguments. How $a^m \equiv -1 \mod p $ and how similar argument works for $b$ but in another case we get inconsistency. | |
| May 4, 2021 at 21:51 | comment | added | Akanksha Gupta | In the paper, we start with the common primitive root $g=xy(mod pq)$. Then how $<x,xy>$ becomes the generating set? | |
| May 4, 2021 at 11:03 | history | edited | LSpice | CC BY-SA 4.0 |
`\mod` almost always spaces badly compared to `\bmod`
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| May 4, 2021 at 8:25 | history | answered | Padraig Ó Catháin | CC BY-SA 4.0 |